feat: add solutions to lc problem: No.0733 (doocs#4087)

No.0733.Flood Fill

Author: yanglbme

solution/0700-0799/0722.Remove Comments/README.md

@@ -59,20 +59,20 @@ tags:
 <strong>解释:</strong> 示例代码可以编排成这样:
 /*Test program */
 int main()
-{ 
-  // variable declaration 
+{
+  // variable declaration
 int a, b, c;
 /* This is a test
-   multiline  
-   comment for 
+   multiline
+   comment for
    testing */
 a = b + c;
 }
 第 1 行和第 6-9 行的字符串 /* 表示块注释。第 4 行的字符串 // 表示行注释。
-编排后: 
+编排后:
 int main()
-{ 
-  
+{
+
 int a, b, c;
 a = b + c;
 }</pre>
@@ -106,15 +106,15 @@ a = b + c;
 
 ### 方法一：分情况讨论
 
-我们用一个变量 $blockComment$ 来表示当前是否处于块注释中，初始时 $blockComment$ 为 `false`；用一个变量 $t$ 来存储当前行的有效字符。
+我们用一个变量 来表示当前是否处于块注释中，初始时 $\textit{blockComment}$ 为 `false`；用一个变量 $t$ 来存储当前行的有效字符。
 
 接下来，遍历每一行，分情况讨论：
 
-如果当前处于块注释中，那么如果当前字符和下一个字符是 `'*/'`，说明块注释结束，我们将 $blockComment$ 置为 `false`，并且跳过这两个字符；否则，我们继续保持块注释状态，不做任何操作；
+如果当前处于块注释中，那么如果当前字符和下一个字符是 `'*/'`，说明块注释结束，我们将 $\textit{blockComment}$ 置为 `false`，并且跳过这两个字符；否则，我们继续保持块注释状态，不做任何操作；
 
-如果当前不处于块注释中，那么如果当前字符和下一个字符是 `'/*'`，说明块注释开始，我们将 $blockComment$ 置为 `true`，并且跳过这两个字符；如果当前字符和下一个字符是 `'//'`，那么说明行注释开始，我们直接退出当前行的遍历；否则，说明当前字符是有效字符，我们将其加入 $t$ 中；
+如果当前不处于块注释中，那么如果当前字符和下一个字符是 `'/*'`，说明块注释开始，我们将 $\textit{blockComment}$ 置为 `true`，并且跳过这两个字符；如果当前字符和下一个字符是 `'//'`，那么说明行注释开始，我们直接退出当前行的遍历；否则，说明当前字符是有效字符，我们将其加入 $t$ 中；
 
-遍历完当前行后，如果 $blockComment$ 为 `false`，并且 $t$ 不为空，说明当前行是有效行，我们将其加入答案数组中，并且清空 $t$。继续遍历下一行。
+遍历完当前行后，如果 $\textit{blockComment}$ 为 `false`，并且 $t$ 不为空，说明当前行是有效行，我们将其加入答案数组中，并且清空 $t$。继续遍历下一行。
 
 时间复杂度 $O(L)$，空间复杂度 $O(L)$，其中 $L$ 是源代码的总长度。
 

solution/0700-0799/0722.Remove Comments/README_EN.md

@@ -58,20 +58,20 @@ tags:
 <strong>Explanation:</strong> The line by line code is visualized as below:
 /*Test program */
 int main()
-{ 
-  // variable declaration 
+{
+  // variable declaration
 int a, b, c;
 /* This is a test
-   multiline  
-   comment for 
+   multiline
+   comment for
    testing */
 a = b + c;
 }
 The string /* denotes a block comment, including line 1 and lines 6-9. The string // denotes line 4 as comments.
 The line by line output code is visualized as below:
 int main()
-{ 
-  
+{
+
 int a, b, c;
 a = b + c;
 }
@@ -102,7 +102,19 @@ a = b + c;
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Case Analysis
+
+We use a variable $\textit{blockComment}$ to indicate whether we are currently in a block comment. Initially, $\textit{blockComment}$ is `false`. We use a variable $t$ to store the valid characters of the current line.
+
+Next, we traverse each line and discuss the following cases:
+
+If we are currently in a block comment, and the current character and the next character are `'*/'`, it means the block comment ends. We set $\textit{blockComment}$ to `false` and skip these two characters. Otherwise, we continue in the block comment state without doing anything.
+
+If we are not currently in a block comment, and the current character and the next character are `'/*'`, it means a block comment starts. We set $\textit{blockComment}$ to `true` and skip these two characters. If the current character and the next character are `'//'`, it means a line comment starts, and we exit the current line traversal. Otherwise, the current character is a valid character, and we add it to $t$.
+
+After traversing the current line, if $\textit{blockComment}$ is `false` and $t$ is not empty, it means the current line is valid. We add it to the answer array and clear $t$. Continue to traverse the next line.
+
+The time complexity is $O(L)$, and the space complexity is $O(L)$, where $L$ is the total length of the source code.
 
 <!-- tabs:start -->
 

solution/0700-0799/0727.Minimum Window Subsequence/README.md

@@ -60,7 +60,7 @@ s1 = "abcdebdde", s2 = "bde"
 
 ### 方法一：动态规划
 
-我们定义 $f[i][j]$ 表示字符串 $s1$ 的前 $i$ 个字符包含字符串 $s2$ 的前 $j$ 个字符时的最短子串的起始位置，如果不存在则为 $0$。
+我们定义 $f[i][j]$ 表示字符串 $\textit{s1}$ 的前 $i$ 个字符包含字符串 $\textit{s2}$ 的前 $j$ 个字符时的最短子串的起始位置，如果不存在则为 $0$。
 
 我们可以得到状态转移方程：
 
@@ -72,9 +72,9 @@ f[i - 1][j], & s1[i-1] \ne s2[j-1]
 \end{cases}
 $$
 
-接下来我们只需要遍历 $s1$，如果 $f[i][n] \gt 0$，则更新最短子串的起始位置和长度。最后返回最短子串即可。
+接下来我们只需要遍历 $\textit{s1}$，如果 $f[i][n] \gt 0$，则更新最短子串的起始位置和长度。最后返回最短子串即可。
 
-时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别为字符串 $s1$ 和 $s2$ 的长度。
+时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别为字符串 $\textit{s1}$ 和 $\textit{s2}$ 的长度。
 
 <!-- tabs:start -->
 

solution/0700-0799/0727.Minimum Window Subsequence/README_EN.md

@@ -28,7 +28,7 @@ tags:
 <pre>
 <strong>Input:</strong> s1 = &quot;abcdebdde&quot;, s2 = &quot;bde&quot;
 <strong>Output:</strong> &quot;bcde&quot;
-<strong>Explanation:</strong> 
+<strong>Explanation:</strong>
 &quot;bcde&quot; is the answer because it occurs before &quot;bdde&quot; which has the same length.
 &quot;deb&quot; is not a smaller window because the elements of s2 in the window must occur in order.
 </pre>
@@ -55,7 +55,23 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Dynamic Programming
+
+We define $f[i][j]$ to represent the starting position of the shortest substring of the first $i$ characters of string $\textit{s1}$ that contains the first $j$ characters of string $\textit{s2}$. If it does not exist, it is $0$.
+
+We can derive the state transition equation:
+
+$$
+f[i][j] = \begin{cases}
+i, & j = 1 \textit{ and } s1[i-1] = s2[j] \\
+f[i - 1][j - 1], & j > 1 \textit{ and } s1[i-1] = s2[j-1] \\
+f[i - 1][j], & s1[i-1] \ne s2[j-1]
+\end{cases}
+$$
+
+Next, we only need to traverse $\textit{s1}$. If $f[i][n] \gt 0$, update the starting position and length of the shortest substring. Finally, return the shortest substring.
+
+The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the lengths of strings $\textit{s1}$ and $\textit{s2}$, respectively.
 
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